Correct.

$\displaystyle{12}{{d}}^{{2}}={2}\cdot{2}\cdot{3}\cdot{d}\cdot{d}\usepackage{cmbright}\Huge\$ and $\displaystyle{18}{{d}}^{{3}}={2}\cdot{3}\cdot{3}\cdot{d}\cdot{d}\cdot{d}\usepackage{cmbright}\Huge\$ , so the LCM $\displaystyle={2}\cdot{2}\cdot{3}\cdot{3}\cdot{d}\cdot{d}\cdot{d}={36}{{d}}^{{3}}\usepackage{cmbright}\Huge\$

$\displaystyle\frac{{{5}{c}}}{{{12}{{d}}^{{2}}}}-\frac{{{5}{c}}}{{{18}{{d}}^{{3}}}}=\frac{{{5}{c}{\left({3}{d}\right)}}}{{{\left({12}{{d}}^{{2}}\right)}{\left({3}{d}\right)}}}-\frac{{{5}{c}{\left({2}\right)}}}{{{18}{{d}}^{{3}}{\left({2}\right)}}}=\frac{{{15}{c}{d}}}{{{36}{{d}}^{{3}}}}-\frac{{{10}{c}}}{{{36}{{d}}^{{3}}}}=\frac{{{15}{c}{d}-{10}{c}}}{{{36}{{d}}^{{3}}}}=\frac{{{5}{c}{\left({3}{d}-{2}\right)}}}{{{36}{{d}}^{{3}}}}\usepackage{cmbright}\Huge\$